Why does the conductance of a solution changes during titration. And what is the relation of conductance of a solution with titration?(for both acid base and redox titration)
2
-
2$\begingroup$ Here is some information for acid/base titrations. You can improve your question by editing it, incorporating what you know and what you need help with. This paper also discusses redox titrations, but it is behind a paywall. However, once you understand the acid/base titrations, you should be able to figure out which redox titrations might work well. $\endgroup$– Karsten ♦Commented Nov 6, 2022 at 17:28
-
1$\begingroup$ I believe this question is missing context about what really is the misunderstanding. What are you confused about? Do you understand why solutions are conductive in the first place? Or is it applying this knowledge to titrations? I think some clarity about what you know and where you’re finding trouble connecting ideas would help us write a good answer. $\endgroup$– Avogadro ♦Commented Nov 6, 2022 at 23:06
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Let's take an example. A solution of hydrochloric acid contains two ions : $\ce{H3O+}$ and $\ce{Cl-}$. The ion $\ce{H3O+}$ has a specially high conductivity at infinite dilution ($\ce{349.8 cm^2ohm^{-1}mol^{-1}}$), about $5$ times as high as the conductivity of $\ce{Cl- (76.4cm^2ohm^{-1}mol^{-1}}$). In a titration with $\ce{NaOH}$, the ion $\ce{H3O+}$ is replaced by the ion $\ce{Na+}$ whose conductivity is low, even lower than $\ce{Cl-}$ : the conductivity of $\ce{Na+}$ is $\ce{50.1 cm^2ohm^{-1}mol^{-1}}$.
So the $\ce{HCl}$ solution has a high conductance, which is mainly due to its "quick" ion $\ce{H3O+}$. If this ion is replaced by the "slow" $\ce{Na+}$, the conductance of the solution decreases.
$\begingroup$
$\endgroup$
Before you post and ask you need to look information on internet because there is a lot of information about this kind of titration.
You must think about what produce the conductivity and what are the changes during redox reactions. In this case theres a change in the free ions and their charges that form the analyte. When you add the titrant you are reacting it with the analyte and makes a stable compund with low conductivity. You assume that the coplete reaction as been done when you reach the equivalent point of conductance. If you still adding titrant you will see an increase of conductivity because of the free ions of this compound.
So when you are looking for the conductance you are looking for the point that make the system stable as you add the reactant that it will be minimum at the equivalent point that means that the complete reaction has been done.
